Question: Divide the following rational expressions and simplify the result. $\dfrac{3k^2n}{n^2+4kn-12k^2}\div\dfrac{9kn^2}{18k^2+9kn+n^2}=$
Solution: Let's first factor the numerators and denominators of each expression separately. [Why are we doing this?] The numerator, $3k^2n$, of the dividend cannot be factored any further. The denominator, $n^2+4kn-12k^2$, of the dividend can be factored as $(n+6k)(n-2k)$ using the sum-product pattern. The numerator, $9kn^2$, of the divisor cannot be factored any further. The denominator, $18k^2+9kn+n^2$, of the divisor can be factored as $(n+6k)(n+3k)$ using the sum-product pattern. Now the division looks as follows: $\dfrac{3k^2n}{(n+6k)(n-2k)}\div\dfrac{9kn^2}{(n+6k)(n+3k)}$ To divide two rational expressions, we flip the divisor, multiply across, then simplify: [What's that?] $\phantom{=} \dfrac{3k^2n}{(n+6k)(n-2k)} \div \dfrac{9kn^2}{(n+6k)(n+3k)} $ $\begin{aligned} &= \dfrac{3k^2n}{(n\!+\!6k)(n\!-\!2k)} \cdot \dfrac{(n\!+\!6k)(n\!+\!3k)}{9kn^2} &\text{Flip the divisor.}\\\\\\\\ &= \dfrac{3k^2n(n+6k)(n+3k)}{9kn^2(n+6k)(n-2k)} &\text{Multiply across.}\\\\\\\\ &= \dfrac{3 \cdot {\cancel{k}} \cdot k \cdot {\cancel{n}} {\cancel{(n+6k)}}(n+3k)}{9 \cdot {\cancel{k}} \cdot {\cancel{n}} \cdot n {\cancel{(n+6k)}}(n-2k)} &\text{Cancel out common factors.}\\\\\\\\ &=\dfrac{k(n+3k)}{3n(n-2k)} \end{aligned}$ Therefore, the simplified form of the quotient is $\dfrac{k(n+3k)}{3n(n-2k)}$, which is equivalent to $\dfrac{kn+3k^2}{3n^2-6kn}$.